Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
vector<int> ke[120000];
int par[17][120000], d[120000];
struct city {
  int a[12];
  city() { memset(a, 60, sizeof(a)); }
  void add(int gt) {
    a[11] = gt;
    sort(a + 1, a + 12);
  }
} val[18][120000];
city xl(city x, city y) {
  city thu = x;
  for (int i = 1; i <= 11; i++) thu.add(y.a[i]);
  return thu;
}
void dfs(int x, int pa) {
  par[0][x] = pa;
  for (int i = 1; i < 17; i++) {
    par[i][x] = par[i - 1][par[i - 1][x]];
    val[i][x] = xl(val[i - 1][x], val[i - 1][par[i - 1][x]]);
  }
  for (int i = 0; i < ke[x].size(); i++) {
    int v = ke[x][i];
    if (v == pa) continue;
    d[v] = d[x] + 1;
    dfs(v, x);
  }
}
int lca(int u, int v) {
  if (d[u] < d[v]) swap(u, v);
  int cl = d[u] - d[v];
  for (int i = 0; i < 17; i++)
    if ((1 << i) & cl) u = par[i][u];
  if (u == v) return u;
  for (int i = 16; i >= 0; i--) {
    if (par[i][u] != par[i][v]) {
      u = par[i][u];
      v = par[i][v];
    }
  }
  return par[0][v];
}
city get(int x, int y) {
  city thu;
  for (int i = 0; i < 17; i++) {
    if ((1 << i) & y) {
      thu = xl(thu, val[i][x]);
      x = par[i][x];
    }
  }
  return thu;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m >> q;
  int u, v;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    ke[u].push_back(v);
    ke[v].push_back(u);
  }
  int c;
  for (int i = 1; i <= m; i++) {
    cin >> c;
    val[0][c].add(i);
  }
  dfs(1, 1);
  int k;
  while (q--) {
    cin >> u >> v >> k;
    int p = lca(u, v);
    city h1 = get(u, d[u] - d[p]);
    city h2 = get(v, d[v] - d[p] + 1);
    city ans = xl(h1, h2);
    int pos = 1;
    while (pos <= k && ans.a[pos] <= m) pos++;
    cout << pos - 1 << " ";
    for (int i = 1; i < pos; i++) cout << ans.a[i] << " ";
    cout << endl;
  }
  return 0;
}